ANALYSIS OF PRESTRi^ooED AND REINFORCED CONCRETE FOLDED PLATES by RAMESH GAMI B. S., Bombay University, 1958 A MASTER'S REPORT submitted in partial fulfillment of the requirements for the degree MASTER OF SCIENCE Department of Civil Engineering KANSAS STATE UNIVERSITY Manhattan, Kansas 1963 Approved by: 'J/.yn, ajor Professor k f ii TABLE OF CONTENTS r>->e.i+s Page SYNOPSIS 1 INTRODUCTION 2 DEFINITIONS 4 STATEMENT OF PROBLEM 6 ANALYSIS OF FOLDED PLATES CONSIDERING JOINT DISPLACE- MENT 7 ANALYSIS 8 Primary Stresses 8 Secondary Stresses 20 ANALYSIS OF POST SSB8X0X5S rRESTREoSED COHCBEH FOLDED PLATE 38 CONCLUSIONS 55 ACKNOWLEDGMENT 57 APPENDIX I - EXPLANATION OF TERMS 58 APPENDIX II - BIBTIOGRAPHT 60 ANALYSIS OF PRESTRESSED AND REINFORCED CONCRETE FOLDED PLATES by RAMESH GAUI 1 SYNOPSIS The simplified methods of analysis for long span pre- stressed and reinforced concrete folded plate structures are presented herein. A reinforced concrete folded plate is analyzed by resolv- ing the applied load into two directions, one vertical and the other parallel to the plate on which it acts. The vertical load will induce "slab" action and the parallel load will in- duce "plate" action. Slab action can be analyzed by assuming non-yielding supports and then applying a correction for de- flection, from which total plate loads are found. "Plate" ac- tion causes a deflection of the longitudinal edge and intro- duces longitudinal stresses. To satisfy the compatibility condition, a stress distribution is done. Moments are com- puted due to relative displacements. From these moments, plate loads and longitudinal stresses are calculated. Graduate student, Department of Civil xungineering, Kansas otate University, Manhattan, Kansas. To analyze the prestressed plate, the stress distribution along the longitudinal axis is analyzed in the same manner as for any homogenous beam of rectangular section. Transverse bending in successive plates is analyzed by the conventional moment distribution method utilized for con- tinuous structures. A typical problem is outlined and the simplified method of analysis is discussed in this report. INTRODUCTION Folded plates, or as they are sometimes called, prismatic shells or hipped plates, provide a useful and economical meth- od of construction for roof and floor systems in a wide variety of structures. They are competitive with other construction methods for short spans and have proven exceptionally economi- cal where relatively large spans are needed as for auditoriums, gymnasiums, industrial buildings, hangers, department stores and parking garages. The folded plate shape of roof structure has come into wide usage because of its low cost of construc- tion for long span, high load carrying capacity, rigidity, and aesthetic interest. Selection of concrete for the shell mate- rial furnishes a high degree of fire resistivity, ease of mold- ing to the desired alinement and profile, a great degree of permanence, and low construction and maintenance costs. Folded plate floor or roof construction consists of a series of repeated units, each of which is formed by tv/o or more flat plates intersecting at an angle. The plates act as a continuous slab transversely and as beams in their own planes. Their structural behaviour resembles that of shells. In fact, a cylindrical shell can be thought of as a folded plate in lim- it. The structural action of a folded plate consists of trans- verse "slab action" by which the loads are carried to the joints and longitudinal "plate action" by which the loads are finally transmitted to the transverses. Because of its great depth and small thickness, each plate offers considerable re- sistance to bending in its own plane. This "plate action" ex- plains the remarkable rigidity of folded plate construction. Folded plates have certain advantages over shells. These advantages are: (1) The shuttering required is relatively simpler as it involves only straight planks. (2) Shuttering can be stripped at the end of seven days, if not earlier, because of their greater rigidity; this results in quicker turnover which, in turn, cuts down construction time. (3) The design involves only simple calculations which do not call for a knowledge of higher mathematics. (4) Movable formwork can be employed for their construc- tion with greater ease than with cylindrical shells. (5) Simple rectangular diaphragms take the place of complicated transverses required for shells. (6) .?heir light reflecting geometry and pleasing out- lines make then comparable with shells in their aes- thetic appeal. Interest in and use of this type of roof has increased considerably in this country. Considerable additions have been made to our analytical and experimental knowledge in the last decade. The purpose of this report is to solve typical problems and discuss procedures which may be employed for the analysis of a single-span folded plate structure. DEFINITIONS The following definitions are used as a basis for the discussion in this report: (1) A plate is an individual planar element of the structure. (2) The length of a plate is the dimension between trans- verse supports. (Fig. 1, "L") (3) The width of a plate is the transverse dimention be- tween longitudinal edges. (Fig. 1, "W") (4) The height of the structure is the vertical dimension of the upper and lower extremes of a transverse cross section. (Fig. 1, "h") FIG. 1. - END PORTION OF A FOLDED PLATE ROOF Support Support FIG. 2. - TYPICAL CROSS SECTION THROUGH MULTIPLE FOLDS STATEMENT OF PROBLEM Consider a simple span of 60 feet in which we shall arbit- rarily assume a column spacing along the ends of this longer span of 23 feet. The slope of each plate will be assumed to be 3 1/2 on 12, vertical to horizontal. (Fig. 2.) A trial analy- sis will indicate this corrugated configuration to be unsuit- able for conventionally reinforced concrete due to the very small area of concrete available at the ridge, causing high compressive stresses and a very high percentage of compressive steel reinforcement. This can be primarily attributed to the shallow ratio of depth from ridge to valley for the long span of 60 feet and the absence of sufficient concrete area in the ridge. Several remedies for this condition become readily appar- ent: (1) Steepen the slope of the plate and add a third plate horizontally at the ridge in order to provide suffi- cient concrete area to reduce the high compressive stresses. (2) Steepen the slope of the plates in order to provide a deeper section from ridge to valley. (3) Provide a combination of each of those stated in (1) and (2). (4-) Prestress the structure in order to utilize all of the concrete area as homogenous section. This report shall advance methods of analysis for design of folded plates by the above mentioned procedures (2) and (4). (a) To steepen the sloj^e of the plates in order to pro- vide a deeper section from ridge to valley. The folded plate is analyzed by considering the effects of relative displace- ments of the longitudinal edges. (b) To prestress the structure in order to utilize all of the concrete area as homogenous section. ANALYSIS OF FOLDED PLATES CONSIDERING JOINT DISPLACEMENT Assumptions The following general assumptions are made in analyzing a folded plate structure: 1. The material is homogenous, uncracked and elastic. 2. Longitudinal edge Joints are fully monolithic and con- tinuous; there is no relative rotation or translation of two adjoining plates at their common boundary. 3. The principal of superposition holds, that is, the structure may be analyzed separately for the effects of its redundants and various external loadings and the results combined algebraically. 4. Individual plates possess negligible torsional resis- tance and torsional stresses due to twisting of the plates can be neglected. 5« The function of the supporting diaphragms or bents is to supply the end reactions for the plate action and 8 for the longitudinal slab action. 1'hey are assumed incapable of providing restraint against rotation of the ends of the plates in their own planes, but may provide some restraint for longitudinal slab bending. 6. Longitudinal strain due to plate action varies linear- ly across the width of each plate (plane section re- mains plane). The rate of change of strain with re- spect to width ordinarily will differ from plate to plate from which it can be inferred that there will be some relative displacement of the joints of a cross section. 7. Longitudinal slab action can be neglected; that is, slab bending carries the load applied to the surface of a plate to the longitudinal edges only, as in a one way slab. ANALYSIS Primary Stresses The folded plate for the structure, as discussed in "Statement of Problem", is deepened at ridge and valley as shown in the line diagram, Fig. 3» FIG. 3. - LIKE DIAGRAM OF FOLDED PLATE STRUCTURES. The data for this sample solution are as follows: Slab thickness = 4 1/2" Span between transverses * . . . 60' 0" Column spacing = 23' 0" Loading: Roofing 5 psf Snow load, insulation, and accoustatics ... 30 psf The loads on the inclined surfaces are as follows: For Plate AB (Fig. 4) Snow Load: FIG. 4. - PLATE AB. 10 cos* £732 = 0.94-9 sin* = 2 0.317 5732 Snow load on inclined surface = 30 x cos* = 30 x 0.949 = 29. 4-7 lbs/sq. in. Dead Load : Hoofing material = 5*00 lbs/sq. in. Slab 4.5 x 150 * 55.33 lbs/sq. in. 12 Total Load : Sum of above three = 29.^7 ♦ 5.00 + 55.33 = 89.80 say 90.00 lbs/sq. in, Components of Total Load : ( Fig . 5 • ) This load can be resolved into normal and tangential components, Wn and Wt. FIG. 5. - NORMAL AND TANGENTIAL COMPONENTS ON PLATE AB. 11 Wn ■ W x cos <x = 90 x 0.949 = 85.41 lbs/sq. in. Wt = W x sin <x = 90 x 0.317 = 28.53 lbs/sq. in. For Plate BC (Jig. 6.) fc-j — FIG. 6. - PLATE BC. Snow Load: Dead Load: « as x 11.5 12.92 ■ 27.62 lbs/sq. in. roofing ■ 5.00 lbs/sq. in. Dead load of slab m 55. 33 lbs/sq. in. Total Load : = 87.95 lbs/sq. ft. Components of Total Load : Normal component Vn ■ 87.95 x 11.5 12.95 = 77.5 lbs/sq. It. Tangential component = Vt = 87.95 x 6 12.92 = 40.60 lbs/sq. ft. 12 The load on DE is by observation the same as the load on £C, that is: w = 77.5 lbs./sq. ft. say 78 lbs./sq. ft. W t = 40.6 lbc/s^. ft. say 41 lbs./sq. ft. For the purpose of analysis, assume fictitious supports at joints B, G, D, and calculate the Lioments and reactions. It is assumed that the relative displacement of joints is not present. A transverse strip one foot wide is considered, treating it as continuous slab supported at the joints by non-yielding sup- ports. The Boment distribution for this condition is performed in Fig* 7* 86#/i 't 76#/ft 8( s# /t t '1 ' 1 ■ \ • ■ ' i f ) ' 1 r ' \ ' ' ' r ^ 1' ,r v | | , V I 1 j ■ v i 1 i> v ■' ' » 1 A i E r * c fD E A E c £ Dist. F. 1 0.5 0.5 1 FEM +1725 -1090 +1090 -1090 +1090 3,0. -635 -317.5 +317.5 +625 final n +1725 -1725 +772.5 -772.5 +1725 -1725 FIG. 7. - SLAB MOMENT DUE TO EXTERNAL LOAD. Reactions are as follows: Total reaction at B ■ 1124 lbs. Total reaction at C s 368 lbs. Total reaction at D = 1124 lbs. 13 These reactions were obtained on the assumption that fic- titious supports exist at B, C % and D and hence, on removal, they give e jual and opposite reactions. These reactions are placed on the plates as shown in fig* 8, FIG. 8. - PLATE SHOWING REACTIONS DUE TO SLAB ACTION. These plate loads due to the above reactions can be ob- tained as follows (see Fig. 9) J tan © x - § - 0.3333 © x « 18° - 26» tan © 2 * jj-r = 0.521 © 2 s 27° - 40' Considering joint B, (see Fig. 10) Comp. along; AB Comp. along BC 54-8 sin(90 + © 1 + 9 2 ) = sin 90 s sin (ISO - © x *P Comp. along AB . ^ 8 x sin <*> » & 1 * V sin (180 - 57 - 9 2 ) = 5^8 x 0.96 = 526 lbs. 14 FIG. 9. - INCLINATION OF PLATES. Comp. along BC ■ 54-8 x sin 90 sin (180 - 57 - 9 2 ) 548 ~ 0.?1 = 770 lbs. 5^8# FIG. 10. - COMPONENTS ALONG AB AND BC DUE TO REACTIONS Similarly due to the load of 576 lbs. the components along AB and BC can be obtained as follows: Comp. alonp; AB Comp. alom BC 576 sin 90 = sin (90 * ^ 4 © 2 ; = sin (180 - 9 X - © 2 ; Comp. along AB = s JPf 1 g Q 3 ^ Q . Qg) * o?fr = 812 lbs# 15 576 x sin (90 f & 1 f © 2 ) Comp. along BC = sin (180 - ^ - © 2 ) = 576 cot (9« + w 2 ) - 576 x 0.96 = 553 lbs. Considering joint C, seo Fig. 11. FIG. 11. - COMPONENTS DUE TC EEACT10H8 A? C. Comp. BC Coinp. along CD sin 54° - 40' " sin 90° Comp. along BC * 434 x ffiB g^L ~ sin 55° ^ x 0.82 = 297 lbs. Comp. along CD = s in 55*- 20 ' 434 " 0.82 » 528 lbs. Due to 434 lbs. reaction Comp. along BC - 528 lbs. Conp. along CD = 297 lbs. 434 sin (180 f 2 9 2 ) 40' "20^ 16 812 316 297 FIG. 12. - TOTAL FORCES AT THE JOINTS The total forces at the joints are shown in Fig. 12. The total load along AB can be calculated as follows Load along plate AB =» 1358 * Wt x 6.32 = 1358 f 29 x 6.32 = 1521 lbs. Similarly total load along plate BC = 2149 f v/t x 12.92 = 2149 f 528 = 2677 Ids. 17 From these loadings we can find longitudinal stresses on each plate. In the analyses one treats each plate as a beam carrying loads and spanning between end diaphragms with no edge shear along joints. Plate AB (Fig. 13.) 1521#/ft 1 1 1 1 1 1 1 1 1 a' so.o' B A rf***- B 1 1880#/in 2 1880#/* FIG. 13. - STRESS DISTRIBUTION ALONG PLATE AB. f 1 M w x L 8 f p w x L x h s x 12 1 8 x 2 x b x h 5 2 T 2 3 x w x L 4 x b x h^ 3 x 1521 x 60 2 x 12 4 x 4.5 x 6.32 x 6.32 x 144 ■ 1880 lbs/sq. in. The stresses in top and bottom fiber are « 1880 lbs/sq. in. T t " t f ju • - 1880 lbs/sq. in. 13 Piatt BC. (Fig. 14.) 2149#/ft B 60.0' V 4.5 m^M FIG. 14. - STRESS DISTRIBUTION ALONG BC. -9ia#/ n a f * . 2 X W Y. If 4 x b x h x 2.49 x 60 x 60 x 12 x 4.5 x 12.92 x 12.^2 912 lbs/sq. in. 1860 ^An 3 - 6 ld&0#/in x 912 #/in FIG. 15. - STRESSES AT JUNCTIONS OF PLATES 19 From the preceding analysis, it Is seen that the stresses at the junctions of plates AB and BC at B are different as shown in Fig. 15» but the compatibility condition requires them to be equal. To satisfy the compatibility condition, a stress distribution similar to the process of moment distribution is done where, Stiffness •< 1_ bh Carry over factor ... = - # Since the thickness of AB and BC are equal, the stiffness fac- tors are as follows: K ab = £732 = 0.160 if L ab 12.92 = 0.078 The stress distribution is as follows: A 'A B % ( Z % D g E DF 0.672 0.328 0.5 0.5 0.328 0.672 Stresses +1880 -1880 -912 +912 +912 -912 -1880 +1880 CO. -325 +650 -318 +159 +159 -318 +650 -325 Final Stress +1555 -1230 -230 +1071 +1071 -1230 -1230 +1555 Sign Convention: Compression + V e Tention - V. e 20 Secondary stresses Assume that plates AB, BC, CD and BE are not rigidly jointed and solve for the free edge deflections. There will be stresses due to rotations of those plates which do not have a free edge. Considering a transverse strip one foot wide at the middle of the structure -5 x L 2 (f. - f K ) S * 48 x E x h where f-. ■ stresses at left edge of plate f. = stresses at right edge of plate. The deflection of plate BC can be computed as follows: f f b S -5 x L 2 (f . - f. ) be 48 x £ x h 5 x 60 x 60 1071 - (-1230) x 144 48 x £ x 12.92 5 x 5600 x 2300 x 144 = 48 x 12.92 x E 18 x 23 x 144 x lQ^ = 48 x 12.92 x E be = E S. , - 96.0jc 10 Similarly for AB g - 5x 60 x 60 (-1230 - 155) x 144 max = 48 x E x 6.32 5 x 3600 x 2785 x 144 ■ ♦ 48 x 6.32 S ab = ^ * 1Q 3 inches. 21 From these individual deflections, the composite deflec- tion is found out "by graphical construction as shown in Figure 17. From Figure 17» we obtained the deflection of 170 x 10"^ BC ■ — i — 6 . This deflection is obtained on the assumption that edges A, B, D, D, and E, are all free but actually this is not the case and the deflection will be very, very small due to the fixity of the Joints. Therefore, moments are calculated which are produced by this relative displacement. In case of a fixed end beam, moments induced at A and B by sinking of sup- ports B will be - — *— , see Fig. 16. FIG. 16. - MOMENT DUE TO DISPLACEMENT, Considering edge C as the fixed end and end B as the end free to rotate, the bending moment at C = • — *r- . 22 s A b " 2 ^¥^ ^ -" s e BC - 9 -H^ »» -i*- e BC : ^^^ ^ units Scale: 1" = 60" FIG. 17. - COMPOSITE DEFLECTION ALONG BC . I = b x t I = 12 1x4 .? 12 5 x 12 1 19 x 12 M. 3 x £ x 170 x 10- where 8 = 19 x 12 x E x 12.92 x 12.92 170 x lp5 E L = 12.92 ft, hi - 1360 lb. ft. The calculation of shears is as follows 23 B 12-92' & 'at© 12.92' J 1360 #•**■ *b K c 12.92 = 105 lbs The components of the loads in the direction parallel to plates are Plate load along AB a ^§j = 148 lbs. Plate load along BC a 105 x 0.96 = 101 lbs. Plate load along BC = 105 x 0.^6 0.82 = 71 lbs. 24 Plate load along BC is therefore = 101 - 71 = 30 lbs. Plate loads along AB and BC are, (see Fig, 18.) Plate load along AB » 148 lbs. Plate load along BG ■ 30 lbs. FIG. 18. - PLATE LOADS ON AB AND BC, From this the moment on plate AB can be obtained as follows: B t E / With tension as (-) and compression as ( + ) R x 60 = 2/3 x 60 x 148 x 30 and a M Q = 2/3 x 148 x 30 x 30 x 5/8 = 55500 ft. lbs. For plate BG the moment is as follows: 30# go' ^L 25 M Q = 2/3 x 30 x 30 x $0 x 5/8 = 75 x 150 = 11250 ft. lbs. M Stress in plate AB » -x „ b x h 2 Z * T — Z « 4 * 3 x 6 »?2 2 x 144 108 x 40 ,3 * 4320 in* Stress in plate AB - ^°? 2 * 12 =» 154 psi Stress in plate BC ■ 8 11250 x 12 Z 7 b x h 2 4.5 x 12.92 x 12.92 x 144 6 Z « 18000 in 5 Stress in plate BC = 112 ^ n ^ n 12 18000 - 7.5 psi Stress condition for plates AB and BC are as shown in Figure 19. 26 1.54- #/in 154- #/i^ ^7-5 #/<rf FIG. 19. - STRESSES IN AB AND BC. Compatibility requires stresses at the junction to be equal and hence stress distribution is carried out as follows: A 3 C 0.672 0.328 -154 +154 -7.5 +7.5 +54.25 -109.5 +53 -26.50 -99.75 +45.5 +45.5 -19.0 cresses CO. Final Check The maximum deflection is obtained once again from which relative deflection is commuted by the graphical procedure as in i ? ig. 21. B ab = -5/48 x L 2 x (f t - f b ) = +5A8 x 60 x 60 (4^5 + 99.7 5) E x 6.32 ■ 12 ' 2 $ * 10? inches be = -5/48 x 60 x 60 (43.5 + 19.0) x 144 12.92 x E 5/48 x 3600 x 64.5 x 144 12#92 - 2 -?° J 1Q ^ inches 27 Individual deflections are as shown in Fig* 20, from which relative deflection at BC can be found out graphically as shown in . 21. Lfta 21, s bc = ll '° | 1Q inches We see that correction applied for a displacement of 1 ? Q 1 1Q causes itself a displacement of 11,Q * 1Q . The 11 x 10^ correction will have to apply for a displacement of *** w A which will be in form of geometric series, such as 2 n a, ar, ar , ar Summation 1 f r In the above case the series will be as shown below after taking out common factor of *jr 170, 170 x £^» 170 x ( n *g^ 170 Summation = , . 11.0 1 * T75 m 170 x 170 " 181 160 P^t^-K Q 160 X 10^ Correct S^ « 1 Actual S, m = Actual Longitudinal stress Actual Moment oc - ■ f Initial S, Initial Longitudinal stress Initial Moment 160.0 = 0.94 28 FIG. 20. - INDIVIDUAL DEFLECTION AT AB AND BC . cj.-d = ^—y. in units r, 2.70 x 10? . b BC = — ' = m units o 11.0 x 10 ?. 6-op = jg in units Scale: 1" = 10" FIG. 21. - RELATIVE DISPLACEMENT ALONG BC . 29 The longitudinal stresses at center line of structure due to deflection are multiplied by 0.94- by which we obtain the actual longitudinal stresses. Longitudinal stresses at the center line of the structure due to deflection are: Initial stresses Final stresses Initial xo 94 -99.75 -95.75 ^5.5 +42.8 45.5 +42.3 -19.0 -17.8 The total longitudinal stresses at center line are: A 1 C +1550.00 -93.75 -1230 +42.80 -1230 +42.80 +1071 -17.80 +1461.25 -1187.20 -1187.20 +1053.20 stresses due to load (psi) Stresses due to de- flections in psi Final stresses in psi Final end moments : A 3 _J * J +1725 -1725 +77-50 -772.50 +1725 +1360 -1360 +1280 -1280 Final end moment due to loading 'inal end moment due to deflection with- out correction Final end moment due to deflection with correction 1360 x 0.94 * 1280 30 Moments at center of long n (60'), Moments due to load w x If c 78 x 12.92 x 12.92 8 = 1620 lb. ft. The moment diagram at center of long span is shown in Fig. 22. FIG. 22. - MOMENT AT CENTER OF LONG SPAN. Moment at quarter point of long span (60') due to deflection (Fig. 23): The moment due to deflection is assumed to have parabolic variation along the span of structure. nT) ,. , Q . 1280 x 13 x 13 Ordinate ab = 30 x 30 — = 32u units Moment at quarter point of long span due to deflection = 1280— 320 = 960 ft. lb. (See Fig. 23.) 31 >^ "TC -fi O .1 u < *' * X5' 15' 1 — ar -w O oO FIG. 23. - MOMENT AT QUARTER POINT DUE TO DEFLECTION. Moment diagram at quarter point: (Fig. 24.) Moment at C * 772 + 1280 - 320 * 1732 ft. lbs. FIG. 24. - MOMENT DIAGRAM AT QUARTER POINT OF LONG SPAN (60') Moment diagram at end of long span (60'): (?ig« 25.) Moment due to deflection ■ FIG. 25 - MOMENT DIAGRAM AT END, 32 Shear force at center line of long span: (Fig. 26.) 86#/ft IU1 ! 1 1 4 1111 1 2* 1725^ ^ 78#/ft B lll 11 1 11 11 i 1 1 1 1 ITT] c 2050#f<t as.ss* FIG. 26. - SHEAR FORCE AT CENTER LINE OF LONG SPAN. For Span BC: Reactions due to Loadings: R B = R c = 78 x 2^2 = 505 lbs. Reactions due to Moment: R . R . 172^ -202? K B - *C 12.92 * - 25.25 lbs. For Span AB: R B = 5^0 lbs. Total shear force at C = 505 - 25.25 ■ 479.75 lbs. B = 505 - 25.25 + 5^0 = 1011.75 lbs. 33 Shear force at quarter point of long span: (Fig. 27 •) 86#/ft a 1 1 1 i i a 1 i i i 1 b 78#/ft <o4-0# ' 505 * 505# FIG. 27. - SHEAR FORCE AT QUARTER POINT OF LONG SPAJN. For sjjan BC: Reactions due to loading R B = R c = 505 lbs. Reaction due to moment ■ negligible For span AB: R^ x 6.32 = 1725 + 86 x 6.32 x —^ R^ = 5-4-0 lbs. Total shear force at G = 505 lbs. B » 505 + 54-0 = 1045 lbs. Shear force at end of long span (60*): (Fig. 26.) 86#/ft BPH 3 ^ilUUUI UUUlj C 772 Ft# : V^ 505.00 * i 54-0* 73-60 #= 73. 6# FIG. 28. - SHEAR FORCE AT EHD. 34 For span BC: Reaction due to loading « 505 lbs, .Reaction due to moment = 1725 - 772 12.92 = 73.6 lbs. Total shear force at G = 505 + 73.6 = 578.6 lbs. B = 5^0 + 578.6 = 1118.6 lbs. Final results may be summarized as follows: Maximum longitudinal stresses (consider compression as positive and tension as negative) : , B G f 1461. 50 psi -1187.20 psi +1053.20 psi Maximum moments (moments in ft. lbs.): B C D Location -1725 -2052 -1725 fe mid span -1725 -1732 -1725 @ 1/4& span -1725 -772 -1725 at end 35 B 6- 32 12-92' t B.M. Diagram is plotted on Tension Side. PLATES AB AND BC. 479.75 SHEAR FORCE AT CENTER OF LONG SPAN. 7 2052 BENDING MOMENT AT CENTER OF LONG SPAN FIG. 29. - BENDiNG MOMENT AND SHEAR FORCE DIAGRAMS AT CENTER OF LONG SPAN. B <t < 6-32' J 12.92' w PLATES AB AND BC . 36 505 SHEAR FORCE AT QUARTER POINT 1732 BENDING MOMENT AT QUARTER POINT. FIG. 30. - SHEAR FORCE AND BENDING MOMENT AT QUARTER POINT B 37 I- 6- W ->*- 12.92' PLATES AB AND BC . 578.6 SHEAR FORCE AT END OF SPAN. 772 BENDING MOMENT AT END OF SPAN. FIG. 31. - BENDING MOMENT AND SHEAR FORCE AT END OF SPAN, t>* . IS OP POSMJiM I0B20 FKEST8E88£I) CONCRETE FOLDED PLATE x'he stress distribution acting axiaily ^long the longitu- dinal axis is the same for any homogeneous beam of rectangular section. The top surface will be in compression, the bottom in tension under gravity loads. Application of a sufficient pre stressing force at a suitable distance below the center of gravity of the section c ; in eliminate the tension stress. The transverse direction of the plate is analyzed as a continuous slab of length equal to the width of the plate and supported at the folded lines of the ridges and valleys. These spans being quite short thus require only nominal steel rein- forcing as dictated by slab thickness and bending moment. Re- ductions of the plate action deflections along the ridges, and valleys, due to application of prestressing force, permit anal- ysis of the transverse slab by moment distribution with a rea- sonable degree of accuracy. Distribution bars are provided in the longitudinal direction in an amount ordinarily utilized for temperature steel. The design of the structure is more easily understood with the assumption of one individual leaf isolated from the others. One may consider the plate oriented in its working position and compute its sectional properties about the horizontal axis "Long Span Prestressed Concrete Folded Plate Roofs", by J. Brough and B. Stephens, Proc. Am. Soc. G. E. , Jan., I960, p. 95. 39 extending through its center of gravity. Figure 32 indicates the terms used in the following presentation. FIG. 32. - SECTION PROPERTIES OF ONE FOLDED PLATE. First a slab thickness is assumed and then the moment of inertia of the cross section for the inclined position to the vertical axis is computed. The moment of inertia of one plate is derived as follows with a slab thickness of 4# inches, nor- mal to the slope, and with the slope of the plate being 3»5 vertical to 12 horizontal. Thus, in Fig. 32: dA - tdx y « tanc£ fe-fc* dx 12 dy -frr ♦H/2 -H/2 ♦H/2 -H/2 y 2 t dx _2+. 12 dv 40 U-l 1^2 fH/2 -H/2 T3~T 12 t " " '■ ' .Li 3 3 12 t / H v T3 ( 57 + ^ H' t ♦H/2 -H/2 H 5 t where t is vertical thickness of the slab i is the slope rise on 12 which is 3.5 H is height of valley to ridge center to center t' is the •!«% thickness normal to slope Thus t - |— .• Ui x P P 12^ + 3.5 * 4,5/0.98 4.6875 in. j . ft^p x (z.o.25) 5 I - 87300 in The tensioning force required for zero stress in the bottom under working load, at midspan is I = e + k. 41 Where F is the prestress force e il the ecentricity of prestress force from the center of gravity. k is the kern point The top fiber stress, at raidspar under workin- load is f FH ct ^ A c°b Where f is the allowable concrete unit stress. c H is the Ytrtieal height of the section valley to ridge A^ is the area of concrete, c The bottom fiber stress at initial condition at tr isfer is f =* F i cb A A c 1 + e - U fx /F ± k t Where subscript i indicates the condition at transfer of pre- stress. Area of one plate = 4.6875 x 11. 5 x 12 = 647 in 2 r A 647 L ^ r r Htt m 40.25 + 4.6875 oo n.n u r 2 k. * k, = •—- = 6.0 in. t b c Where k. is top kern distance k. is bottom kern distance r is the radius of gyration 42 If one assumes the center of gravity of the tendons as being e' above bottom of section at midspan, the prestress force eccentric arm is e ■ c - e* = 16.4-7 i'-i. Assume e' ■ 6.0 in. The longitudinal bending moments acting on the section for the dead load of the slab, the applied dead loads such as the roofing, insulation and ceiling, and the live load for which the structure shall be designed are computed as follows: Dead load for slab &*S 12 144 " ^ K? on inclined surface Vertical load 2Z— a ZS- at S6 PS cos<$> 079^ jK> '^ Total load per plate 11.25 x 56.25 = 646 PSF. Additional applied dead load: Roofing (five ply built up asphalt and felt) * 5*0 FSF Concrete insulation and }4 in. acoustical plaster = 5»0 PSF Total applied dead load: 10 x cos <J> ■ 9.6 Applied dead load per plate 9.5 x 11.5 = HI lbs. Snow load per plate 25 x 11.5 ■ 287.5 lbs. 43 Moments can be computed as follows: M^ slab - 0.6*6 X *jp • 291*0 ft-kip 2 M dl a PP lied - 0.111 x ^|- * 50.0 ft-kip 2 HL, * .2875 x S|- « 129.6 ft-kip M tl " 47 °* 6 ft ~ ki P The required prestre3sing force to furnish zero stress in the bottom -\t mid span with total load moment is M ti 470.6 x 12000 , n ft , . a 88 16.47 + 6.0 = 250 '° klps ■ e + K t 16.47 F o " $k * ** kipS P * Total effective prestress after deducting losses. F ■ Total prestress just after transfer, using say 4 cables, requires F = 250/4 «= 62.5 kips per cable. F. ■ 73^5 kips per tendon. Initial tensioning force required for each cable should increase adequately for overcoming tendon friction and wobble of conduit. F H Unit stress at mid span = F. = -r—g c b 250 x 44.94 x 1000 = 647 x 22.47 = 770 vs i top comp. = psi bottom comp. 44 Under the slab load only, the unit stresses at time of transfer or prestress force will be i • - K ¥ 16.47 - 291. J 57o x 2.54 x 1000 = 1155 psi comp. Using allowable steel and concrete stresses specified by the Joint Committee 323 ACI, ASCE for pres tressed concrete for 2 tensioned member. Prestressing steel f 's « 240,000 psi f 's yield = 210,000 psi s ultimate strength of steel f ' at design load ■ 3750 psi c determine the area of prestress steel A s = f~ s lffioOO * *»73* ■*• in "» sa y 1 -75 sq. in. s Check the unit shear stress: V = hi | fiS « 33.3 k ips Assume tendons placed in parabolic curve with c-k = e » 22.47 - 6.0 - 16.47 in above bottom at ends Joint Committee $23 ACI, AJ3CS, for crestrsssed concrete for tensioned members. 45 FIG. 33. - LOCATION OF FRESTRESS FORCE ARRANGED IN PARABOLIC CURVE. Shear at cracking load in steel: xr 4 x F x h V s ' L Where V is shear in steel s F is prestressing force after losses L is length of span, center to center of supports; 61.5' h is vertical distance below center of gravity of con- crete section, 10.45 in. v 4 x 286 x 10.45 V s * 61.5 x 12 * 142 kips Shear force in concrete: V Q v„ = c^ C f Where Q is statical moment about center of gravity of concrete section 46 I is moment of inertia of section b is horizontal projection of i>late width 19.1 x 23 60 loc; mm4 v c = 87300 x 43 " 125 pSi The principal tensile 3 tress is gi;-en by s t v c 2 ♦ (t c /2f- - f 0/2 Where f is allowable concrete unit stress c f c ' ^T 1 = 386 psi s t = ( >5 2 ♦ 193 2 « 37 psi, which is satisfactory, since the allowable tensile stress is 110 psi. Check shear at ultimate load; u 13- Where V is ultimate concrete shear c 2V - v g = 66.6 - 14.2 * 52.4 kips v is ultimate shear stress 52.4 x 2560 v u " 87300 x 4.5 * 351 psi 46 a Ultimate principal tensile stress is given by V v u 2 +< f c/2> 2 * f c/2 551 2 +193 2 - 193 » 199 psi, which is satisfactory, since the allowable tensile stress is 300 psi, thus design is satisfac- tory. Check section for cracking moment: The factor of safety against cracking moment * M/M g Where M - M Q + M„ s c M 8 = F (e + k t ) 286 (16.47 ♦ 6.0) 12 = 514.10 ft. kip. M F_.I 0.14 f • x I r^ ■ c * c c % « c = °- 14 f 2 ^° x x il 7?0 ° ■ w "• "*• The factor of safety against cracking: 514.10 - 170 , y. 525 i# ^ The factor of safety against live load cracking: M - M ; dl M 11 47 Where H • M. ♦ ML ■ 684.10 ft. kip. s c ^dl * s total dea( * load moment » 341.0 ft. kip. M,, is live load moment . . . - 129.6 ft. kip. The factor of safety against live load cracking: 684.10 - 341.0 = 120.6 — = 2.65 The factor of safety considering ultimate moment: The factor of safety for total load "u ^n~ ~~ -m ultimate tensile force x ultimate lever arm Where fi * pj s Ultimate tensile force ■ f ' . A a c s * 240,000 x 1.75 x 0.001 m 420 kips. Ultimate lever arm can be computed as below: (See Fig. 34.) Ultimate lever arm m 44.94 - 6.0 - distance of center of concrete resistance below top Center of concrete resistance if found as below: (See Fig. 34-.) Equivalent width of top section « 12 * 4 *5 = 15.4-3 in. Area of top triangle = ^* 6< ? * 1 ^' 4 ? « 36.2 sq.in. 48 FIO. 34. - ULTIMATE MOMENT STRESS CONDITION. Additional compressive area of stem necessary is m Total area under compression - area of top triangle » 420 x 1000 5190 * 95. 3 s<i« in - 36.2 Additional vertical depth of stem required is ■ fill? = 6 - 2 in - Taking moment about top for finding center of concrete resistance "C ''. cr c m ?6.2 x 2/3 x 4.6? jfcjj (6.2/2+4.6? ) ^cr 131.5 ■ b.5 in. below top Ultimate lever arm is « 44.94 - 6.0 - 6.5 * 32.44 in. Ultimate moment - " 20 i^ 2 - m . 113 ft. kip. 49 The factor of safety for total load M s 11^8 m O The factor of safety for live lord - M ; M di 1138 - 341 * 129.6 = 6.15 Transverse bending in successive plates is analyzed by the conventional moment distribution method utilized for con- tinuous structures. The valley and ridge serve as support for the slab. This being a short spsji, 11" - 6" f moments will be no greater than would be encountered in a level slab with sup- ports at the same frequency. Analysis of a 1 ft. wide strip transverse to the main span: W dl * 54 ♦ 10 = 64 lbs. w\ • » 25 cos (J) = 24.5 lbs. V total load =64+24.5 = 88.5 lbs. Assuming a strip of unit width of cross sectior of the structure to act as a continuous one way slab on unyielding 50 supports, the ridge moments are determined by the moment dis- tribution. Because of symmetry, the slab may be assumed as rigidly fixed at ridge. 11.5* A B 1 5/7 V7 +1725 -1060 -665 +1060 -333 -1060 +1060 +143 +190 +95 +1725 -1725 +870 -870 +1155 FIG. 35. - RIDGE MOMENTS. Ridge moments are computed in Figure 35* Maximum bending moment at the center of span occurs when that span and every other alternate span is loaded. Thus, bend- ing moment at center due to uniformly distributed loading can be ^"Design of Folded Plates", by Eliahu Tranm, Proc. Am. Soc. C.E., Vol. 85, Oct., 1959, p. 87. 51 obtained by the following formula: riidsj;an positive bending moment ■ rr- (w, + V, /p) L = }g x (64 ♦ 12.25) x 12 2 = 912 lb. ft. Distribution bars are i>rovided in the longitudinal direction in an amount ordinarily utilized for temperature steel. The positive moment reinforcement, extending transversely, was placed on top of the distribution steel. The prestressing ten- dons were placed directly on top of the bottom steel and tied to the bottom mat at the correct location. Lfter all tendons were placed, the end bearing plates were substantially anchored to the end form. Reinforcing bars for slab negative moments at valleys and ridges, stirrups, ties and grids for resistance of bursting of concrete at the end anchorage were placed last. Deflection: The moment due to pres tress is P x e = 294 x jo ■ W kip-ft for uniform loading plus 294 x p ' * 256 kip-ft for parabolic prestress. K dl slab = 29.10 ft-kip M dl a P^ lied ■ 50.0 kip-ft. n i;L = 150.6 ft-kip. 52 Upward deflection due to uniform prestress of 14-7 kip-ft: o W x IT 147 x 6Q 2 x 12 2 x 12000 8 x 3,500,000 x 87300 0.374 in. Upward deflection due to parabolic prestress is 5 x V x L 2 * Sth — 5 x ?56 x 60 2 x 12 2 x 12000 = 3,500,000 x 87300 « 0.5^3 in. Total instantaneous upward deflection due to prestress - 0.374 + 0.543 * 0.917 in. Deflection after losses » 0.85 x 0.917 • 0.78 in. Downward deflection due to slab load 5 x 291 x 60 2 x 12 2 x 12000 48 x 3,500,000 x 87300 = 0.617 in. H«n04 immediate upward deflection at transfer is 0.917 - 0.617 - 0.300 53 Dead load deflection downward 30 x 0.617 * 391.0 « 0.162 in. Net downward deflection after losses of prestress and effect of creep is » 0.300 - 0.162 « 0.138 in. Instantaneous downward deflection due to live load moment is 130.6 x 0*617 291.0 - 0.275 in. The resulting camber due to application of prestressing force supports the plate in its geometrical position thus eliminating critical secondary stresses caused by rotation of the ends at the supports that would be inevitable with deflections encoun- tered in conventionally reinforced concrete. The ends of the plates behind the anchorage are thickened to provide for distribution of the concentrated force applied by the tendoms. Additional reinforcing in the form of trans- verse and tie steel is provided in the area immediately behind the tendon anchorages to resist the bursting of the concrete A, created by the ores tress force. The end leaves of a transverse section through the 4 "Long span prestressed concrete folded platf* roofs", by J. Brough and B. Stephens, Proc. km. 3oc. 0. *•« Jan. I960, Vol. 85, p. 91. 5* structure must be provided with tension ties to resist the horizontal force component. This may be accomplished by pro- viding a tie beam, gable wall or other means, at or near the support, that is adequate for resistance of horizontal reac- tion. 55 CONCLUSIONS I'he first structures of this kind were large coal bunkers designed and erected by G. Ehlers of Germany in 1924—25. The first paper on the subject was published by him in 1930. His analysis assumed the longitudinal joints to be hinged, neglect- ing the transverse moments at the junction of the plates. The displacement of joints was also ignored. This theory was im- proved upon in 1932 by £• Gruber who included the effects of transverse continuity and joint displacements. Assuming the joints to be hinged as a first approximation, he developed a solution in the form of simultaneous differential equations of the fourth order, which were solved by the use of rapidly con- verging series. This approach involves (7n + 2) unknowns for (n + 1) plates; thus, a roof of 5 plates would involve 30 un- knowns. Although solution proposed by Gruber was very labori- ous, his conclusion that the assumption of hinged or rigid joints would considerably affect the final results was signifi- cant. This work was followed by that of Cramer who published a paper in 1953. He laid down rough limits in terms of the length bo width ratio of individual plates for their classifi- cation as "long" and "short". The paper by Winter and Pei pub- lished in 194-7 is a landmark in the theory on che subject as thi,/, for the first time, reduced the algebraic solution into a stress distribution procedure analogous to the well-known mo- ment distribution method. However, they neglected the displace* ment of the joints. For short folded plates their approach 56 offers a very simple desig,_i procedure. However, for long plates the joint deflections cannot be ignored, Girkmann, in his book published in 194-8, takes into account joint displace- ments. Treating transverse moments at the joints as the un- knowns, he formulated conditions for the compatibility of lon- gitudinal stresses and displacements at joints. The method leads to as many simultaneous equations as the unknown trans- verse moments. The paper presented by Whitney at the joint ASCE-IABSE meeting in New York is a presentation in English of the Girkmann method with some modification. Gaafar in 1953 published a modification of the Winter and Pie method extended to include the effect of joint displacements. Among available methods, the winter and i ; ie procedure is the simplest. It is applicable only to short folded plates for which the joint displacements can be ignored without appreciable error. Of the methods that are applicable to folded plates of all proportions, those due to Gaafar which is considered in this report seems to be most suitable. 57 ACKNOWLEDGMENT The writer wishes to express his sincere gratitude to Dr. John Mc Entyre for his kind guidance and assistance in the preparation of this report. 58 APPENDIX I - EXPLANATION OF TERMS a ■ Slope rise on 12 A = Area of concrete c A ■ Area of steel e.g. = Center of gravity dl ■ Dead load e = Ecentricity of prestress force from the e.g. e' = Distance from surface to center of prestress force F ■ Prestress force after loboes F ■ Modulus of rupture of concrete F^ = Prestress force at tiiae of transfer o f = Allowable concrete unit stress c f ■ ■ Design concrete strength f ■ Allowable unit stress s H ■ Vertical height of section valley to ridge h * Vertical distance above or below e.g. of concrete section I ■ Moment of inertia of section i ■ Subscript indicating condition at transfer of prestress k = Kern point 11 = Live load M m Bending moment Q ■ Statical moment about e.g. r = Radius of gyration S = Principal tensile stress S' = Principal tensile stress for ultimate 59 S. * Principal tensile stress S. , » Ultimate principal tensile stress c t = Thiclmess of slab normal to section t' * Thickness of slab vertically tl - Total load v = Unit 3 hear v » Unit shear stress in concrete c v = Ultimata shear stress V « Ultimate concrete shear c 60 APPENDIX II - BIBLIOGRAPHY "Design of Folded Plate Roofs," by Howard Simpson, Proc. Am. Soc. C. E. , Vol. 84, No. STI, January, 1958. "Hipped Plate Construction," by G. Winter and M. Pie, Jour. Am. Con. Inst., Proc. Vol. 43, January, 1947. "Reinforced Concrete Folded Plates Construction," by G. S. Whitney, B« G. Anderson, and N. Birnbaum, Jour. Struct. Div. , Proc. Am. Soc. C. £•, Vol. 84, January, 1958. "Hipped Plate Analysis, Considering Joint Displacements," by I. Gaafar, Tran. Am. Soc. C. £. , Vol. 119, 1954. "Design of Folded Plates," by Eliahu Traum, Proc. Am. Soc. C. X., Vol. 85, No. STI. 8., October, 1959. "Long Span Prestressed Concrete Folded Plate Roofs," by J. Brough and B. Stephens, Proc. Am. Soc. C. E. , Vol. 119, 1954. "Design of Prestressed Concrete Structures," by T. Y. Lin. "Tentative Recommendations for Prestresaed Concrete," by ACI-ASCE Joint Committee 325. "Design and Calculation of Reinforced Concrete," by K. L. Rao. "The Analysis and Design of Folded Plates," by Ramswamy, Ramaiah and Jain, Indian Concrete Journal, Jul} , 1961. ANALYSIS OF PRESTRESSED AND REINFORCED CONCRETE FOLDED PLATES by RAMESH GAKI B. S., Bombay University, 1958 AN ABSTRACT OF A MASTER'S REPORT submitted in partial fulfillment of the requirements for the degree MASTER OF SCIENCE Department of Civil Engineering KANSAS STATE UNIVERSITY Manhattan, Kansas 1963 Approved by: Mix* ajor Professor The simplified methods of analysis for lor.p. spun pro- stressed and reinforced concrete folded plr>te structures are presented herein, A reinforced concrete folded plate is analyzed b,/ solving the applied load into two directions, one vertical and the other parallel to the plate on which it acts. Vertical load will induce "slab" action and the parallel load will induce "plate" action, olab action can be analyzed by assuming non- yielding supports and then applying a correction for deflec- tion, from which total loads are found, 1 ,e" action causes a deflection of the longitudinal edge and introduces longitu- dinal stresr.es. To satisfy the compatibility condition a stress distribution is done, homents are computed due to rela- tive displacements. Prom these moments plate loads and longi- tudinal stresses are calculated. This process is repeatedly carried out and eve y time correction is applied. This correc- tion will be in form of geometric series, 3ummation of this series gives the actual deflection. The ratio of actual de- flection and initial deflection gives the multiplier which, when multiplied with initial lon ; ituainal stresses, ^ives ac- tual longitudinal stresses. The final longitudinal stress is calculated by adding stresses due to loading and stresses due to deflections. tmantl are calculated at quarter point, end, and at half distance of the longitudinal edge. To analyze the prestressed plate, the stress distribution along the longitudinal axis i3 analyzed in bhe same manner as for any homogenous beam of rectangular section. Transverse bending in successive plates is analyzed by the conventional moment distribution method utilized for con- tinuous structures. A typical problem is outlined and the simplified method o- analysis is discussed in this report.